22 This result has two fundamental limitations The convergence result involves a condition on G. But often interconnections depend on the states of the agents. What about state-dependent graphs? under investigation see Bullo et al., Aeyels/De Smet, Blondel/Hendrickx The global convergence argument does not extend to nonconvex spaces like the circle, sphere,... How do synchronization algorithms behave globally on manifolds? topic of this talk

23 An algorithm with the same local behavior can be designed on the circle

24 In the following we will extend this to other perfectly symmetric nonlinear spaces = compact homogeneous manifolds (CCH) Formally : quotient manifold of a Lie group by a subgroup Intuitively: all points are identical Examples: sphere S n rotation matrices SO(n) (and all other compact groups) Grassmann manifolds (see last part) In this talk:

25 An alternative distance measure yields convenient properties

26 An alternative distance measure yields convenient properties

27 An alternative distance measure yields convenient properties

28 The induced arithmetic mean of the chordal distance is easily computable

29 The induced arithmetic mean of the chordal distance is easily computable

34 Synchronization is ensured locally. The global behavior is a priori unclear.

35 Fixed but directed graphs can lead to limit cycles, quasi-periodic behavior,...

36 Undirected graphs ensure convergence to an equilibrium set, but which one? What about repulsive agents? - on R n - on circle Agents drive away to infinity Stable equilibria are not trivial

37 The existence of local equilibria is sensitive to the attraction profile between connected agents

38 Alternative algorithms can overcome spurious local equilibria of standard consensus motion Gossip algorithm = forced asynchrony At each time, select a single link, and only its 2 agents move towards each other Thm: If G is uniformly connected, synchronizes with probability 1 also on the circle, sphere,...

39 Alternative algorithms can overcome spurious local equilibria of standard consensus motion Auxiliary variables (can be written with agent-based coordinates)

50 Two particular sets in P k-1 R So = all states belong to a discrete set of k orthogonal lines = colors Sp(G) = every edge is stretched to the maximum distance Properties : Sp(G) can be empty depending on G Sp(G) global maxima of W if (*) So Sp(G) if and only if G is k-colorable (**)

51 The complexity result Question : Given G(V,E) and P k-1 R, is any point in So a stable equilibrium for the repulsive agents? yes/no question (typical decision problem) about specific property of equilibrium set Theorem: This question is as difficult as graph coloring -- that is NP-hard for k>2 -- if g(x) satisfies

56 Can we use the distributed dynamical system to solve graph coloring? Stable equilibria = local maxima of W result about complexity of characterizing stable equilibrium set (as complex as deciding graph coloring) OK possibility to solve graph coloring by swarm optimization? (continuous evolution of the swarm converges to solution = distributed analog computation)??

57 NO The multi-agent system on P k-1 R does not solve graph-coloring Global maxima of W in So Sp(G) graph k-coloring Sp(G) always stable ensure unstable So Multi-agent system for colorable G converges to Sp(G) So Sp(G) [Kochen-Specker Theorem] There exist non-colorable G for k=3 with Sp(G) A system that converges to a point in Sp(G)\So can correspond to colorable or non-colorable G...

58 The Kochen-Specker theorem discusses fundamentals of quantum measurement element of P k-1 R possible result of projective quantum measurement on R k Kochen-Specker: For k 3, there does not exist a function f from the set of possible measurement projectors Pi P k-1 R to associated measurement results in {0,1} such that for every {Pi} that form a physical observable (i.e. that commute and Pi = I) we have f (Pi) = 1 Use: show a contradiction with classical re-interpretations of quantum laws

59 The Kochen-Specker theorem discusses fundamentals of quantum measurement Proof: Constructs an example of N elements of P k-1 R, where mutually orthogonal lines are connected in a graph. Then assigning f (color 1)=1, f (other colors)=0 would solve the task if colorable They have a counterexample with N=31 agents for k=3 They construct a situation where all pairs of connected agents are orthogonal in R 3, but the graph is NOT 3-colorable

61 Conclusion Graph-coloring complexity of consensus on projective space For a class of repulsion functions (robustly difficult) Link not bi-directional: provides no solution for graph coloring Equilibrium stability as key feature to characterize NP-hard for k>2 leaves open the case k=2 correspoding to the circle (which seems not trivial, but further unclear how hard)

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